Question

Two capacitors 2μF and 4μF are connected in parallel. A third capacitor of 6 μF capacity is connected in series. The combination is then connected across a 12V battery. The voltage across 2μF capacity is

• A. 2V
• B. 6V
• C. 8V
• D. 1V

Answer 1

Answer: (B)

6V

Answer 2

Resultant capacitance of condensers of capacity 2μF and 4μF when connected in parallel.

$C' = 2 + 4 = 6$μF

This is connected in series with a capacitor of capacity 6μF in series. The resultant capacity C is given by

$\frac{1}{C} = \frac{1}{6} + \frac{1}{6} = \frac{1}{3}$or$C = 3$μF

Charge on combination q = (3 x 10^-6) x (12) = 36 x 10^-

⁶C

Let the charge on 2μF capacitor be $q_{1}$, then

$\frac{q_{1}}{2} = \frac{q - q_{1}}{4}$or $q_{1} = \frac{q}{3}$

$\therefore q_{1} =$12 x 10^-6C

Now potential across 2μF condenser

=$\frac{q_{1}}{2\text{ x 1}\text{0}^{\text{-6}}} = \frac{12\text{ x 1}\text{0}^{\text{-6}}}{2\text{ x 1}\text{0}^{\text{-6}}} =$6V